Transcript Chi Square Analysis

UNIT 6: MENDELIAN GENETICS
CHI SQUARE ANALYSIS
Ms. Gaynor
AP BIOLOGY
CHI SQUARE ANALYSIS
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The chi square analysis allows you to use
statistics to determine if your data is
“good” or not.
allows us to test for deviations of observed
frequencies from expected frequencies.
Lab #8 = working with Nasonia (jewel
wasps) and Chi Square Analysis
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The following formula is used
You need 2 different hypotheses:
1. NULL Hypothesis
•Data are occurring by chance and it is all
RANDOM!
2. Alternative Hypothesis
•Data are occurring by someoutlside
force. It is NOT by chance and it is NOT
RANDOM!
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This statistical test is compared to a theoretical
probability distribution
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These probability (p) values are on the Chi Square
distribution table
HOW DO YOU USE THIS TABLE
PROPERLY?
you need to determine the degrees of freedom
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Degrees of freedom is the # of groups
(categories) in your data minus one
(1)
If the level of significance read from the table
is greater than .05 or 5% then your
hypothesis is accepted and the data is useful
Two Types of Hypotheses:
1. NULL HYPOTHESIS
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states that there is no substantial
statistical deviation between observed
and expected data.
 a hypothesis of no difference (or no effect)
is called a null hypothesis symbolized H0
In other words, the results are
totally random and occurred by
chance alone.
The null hypothesis states that the two
variables are independent, or that there is NO
relationship to one another.
Two Types of Hypotheses:
2. ALTERNATIVE HYPOTHESIS
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states that there IS a substantial
statistical deviation between
observed and expected data.
 a hypothesis of difference (or effect) is
called a alternative hypothesis
symbolized H1
In other words, the results are
affected by an outside force
and are NOT random and di NOT
occur by chance alone.
Null Hypothesis Example
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A scientist studying bees and butterflies.
Her hypothesis was that a single bee visiting a
flower will pollinate with a higher efficiency than a
single butterfly, which will help produce a greater
number of seeds in the flower bean pod.
We will call this hypothesis H1 or an alternate
hypothesis because it is an alternative to the null
hypothesis.
What is the null hypothesis?
 H0: There is no difference between bees and
butterflies in the number of seeds produced by
the flowers they pollinate.
2 Types of Chi Square Problems
1. Non-genetic Chi Square:
Null Hypothesis:
 Data is due to chance and is completely
random. There is no preference
between the groups/categories.
 Alternative Hypothesis
 Data is NOT due to chance and there IS
a preference between group/categories.
Data is not random.
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2 Types of Chi Square Problems
2. Genetic Chi Square
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Null Hypothesis:
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Punnett Square possibilities will be seen!
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Segregation and independent assortment ARE
occurring (which occur RANDOMLY)
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There is no gene linkage (only IF 2+ genes are involved
in test!)
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Data is NOT significant!!!
Alternative Hypothesis
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Punnett Square possibilities are NOT seen
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Segregation and independent assortment are NOT
occurring
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Gene linkage could be occuring (only IF 2+ genes are
involved in test!)
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Data IS significant!!
Let’s look at a fruit fly
cross and their phenotypes
x
Black body, eyeless
Wild type
(brown body, red eyes)
(bbee)
F1: all wild type
(BbEe)
(BBEE)
F1 x F1
5610
Wild type
1881
Eyeless, Wild type (brown eyes)
622
Black body, eyeless
1896
wt (red eyes), Black body
Analysis of the results
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Once the numbers are in, you have to
determine the expected value of this cross.
This is your hypothesis called the null
hypothesis (no gene linkage is occurring
btw the 2 genes).
What are the expected outcomes of this
cross?
F1 Cross: BbEe
x
BbEe
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9/16 should be wild type (D, D)
3/16 should be normal body eyeless (D, R)
3/16 should be black body wild eyes (R, D)
1/16 should be black body eyeless (R, R)
Now Conduct the Analysis:
To compute the hypothesis value
take 10009/16 = 626
(a.k.a- 1/16 of total offspring)
Now Conduct the Analysis:
Remember: To compute the
hypothesis value take 10009/16 = 626
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1.
Using the chi square formula compute the chi
square value (χ2) for this cross:
Calculate (o-e)2/ e for EACH phenotype
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2.
Sum all numbers to get your chi square value
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(5610 - 5630)2/ 5630 = .07
(1881 - 1877)2/ 1877 = .01
(1896 - 1877 )2/ 1877 = .20
(622 - 626) 2/ 626 = .02
 2= .30
Determine how many degrees of freedom are in
your experiment
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4 (phenotype) groups– 1 = 3
I Have my Chi Square
Value (X2)….What next?
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Figure out which hypothesis is accepted:
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your NULL hypothesis= 9:3:3:1 ratio is seen due
to independent assortment/ segregation is
occuring. Genes are NOT linked.
The alternative hypothesis = any change from
the expected is due to SOME OUTSIDE FORCE!
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IT IS NOT RANDOM!  THE GENES COULD
BELINKED!
To figure which hypothesis is accepted, you
need to use the CHI SQUARE TABLE, which
list CRITICAL VALUES!
CHI SQUARE TABLE
CHI-SQUARE DISTRIBUTION TABLE
Accept Null Hypothesis
(chance ONLY…Data is NOT significant)
Reject Null
Hypothesis
(NOT chance
ONLY)
Probability (p)
Degrees of
Freedom
0.95
0.90
0.80
0.70
0.50
0.30
0.20
0.10
0.05
0.01
0.001
1
0.004
0.02
0.06
0.15
0.46
1.07
1.64
2.71
3.84
6.64
10.83
2
0.10
0.21
0.45
0.71
1.39
2.41
3.22
4.60
5.99
9.21
13.82
3
0.35
0.58
1.01
1.42
2.37
3.66
4.64
6.25
7.82
11.34
16.27
4
0.71
1.06
1.65
2.20
3.36
4.88
5.99
7.78
9.49
13.38
18.47
5
1.14
1.61
2.34
3.00
4.35
6.06
7.29
9.24
11.07
15.09
20.52
6
1.63
2.20
3.07
3.83
5.35
7.23
8.56
10.64
12.59
16.81
22.46
7
2.17
2.83
3.82
4.67
6.35
8.38
9.80
12.02
14.07
18.48
24.32
8
2.73
3.49
4.59
5.53
7.34
9.52
11.03
13.36
15.51
20.09
26.12
9
3.32
4.17
5.38
6.39
8.34
10.66
12.24
14.68
16.92
21.67
27.88
10
3.94
4.86
6.18
7.27
9.34
11.78
13.44
15.99
18.31
23.21
29.59
P
value=
data is
due to
chance
ALONE
In biological applications, a probability 5% is usually adopted as
the standard conventional criteria for probability to have statistical
significance is 0.001-0.05
Is My Data Significant?
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DATA IS SIGNIFICANT IF…
 Null hypothesis is rejected
 p value is equal to 5% (0.05) or less
 This means that there is a 5% or less
chance that the data is due to chance
(randomness) ALONE
 There is a 95%+ chance that data is due
to some OUTSIDE FORCE
 Examples of outside forces
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miscounting, gene linkage, preference,
availability of subjects, etc
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Looking statistical values up on the chi square
distribution table tells us the following:
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the PROBABILITY (P) value read off the
table places our chi square value of 0.30
closer to .95 or 95% (~94%)
This value means that there is a 6% chance that
our results are biased and due to gene linkage.
In other words, the probability of getting our
results is 94%.
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94% of the time when our observed data is this close to
our expected data, this deviation is due to random
chance.
We therefore accept our null hypothesis.
PRACTICE PROBLEMS
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What is the critical value at which we
would reject the null hypothesis for the
fruit fly example earlier?
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For 3 degrees of freedom the value for
our chi square must be > 7.815 to accept
the alternative hypothesis and support
that gene linkage is occurring.
What if our chi square value was 8.0 with 4
degrees of freedom, do we accept or reject
the null hypothesis?
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Accept, since the critical value is >9.48
with 4 degrees of freedom.